Problem: Let $f(x)=\dfrac{x^2}{\ln(x)}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2x\ln(x)+x$ (Choice B) B $2x-\dfrac1x$ (Choice C) C $\dfrac{2x\ln(x)-x}{(\ln(x))^2}$ (Choice D) D $2x^2$
Solution: $f(x)$ is the quotient of two, more basic, expressions: $x^2$ and $\ln(x)$. Therefore, the derivative of $f$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2}{\ln(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(x^2)\ln(x)-x^2\dfrac{d}{dx}(\ln(x))}{(\ln(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{2x\cdot \ln(x)-x^2\cdot \dfrac1x}{(\ln(x))^2}&&\gray{\text{Differentiate }x^2\text{ and }\ln(x)} \\\\ &=\dfrac{2x\ln(x)-x}{(\ln(x))^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=\dfrac{2x\ln(x)-x}{(\ln(x))^2}$